Fractional Exponential Indice: Working Out and Explanation | Step-by-Step Guide

[Sirsh]

I have been given a Indice. I've been trying to figure it out for awhile and need some assistance, It'd be great if someone could work it out and show me the steps they did and even explain it.

([a^-2b^-3/ 2a³b^-4]² )/ [ab^-15]/a^-3b²])


I've attempted the working out in my book, Its abit hard to put the whole working on here as it's time consuming. I'm down to [4a^2b^2]/[a^-12b]

thank you.

 

[Mark44]

Your expression is equal to
$$\left[{\frac{a^{-2}b^{-3}}{2a^3b^{-4}}\right]^2~\frac{a^{-3}b^2}{ab^{-15}}$$

One approach would be to move all the factors with negative exponents from the top to the bottom or from the bottom to the top in each fraction, which would leave you with positive exponents. The idea is that a^-n = 1/aⁿ. After that, square each factor in the first fraction. Finally, cancel everything that can be cancelled.

 

[Sirsh]

Thanks for that :) is this how i do it?

[a^-2b^-3 / 2a³b^-4 ]² / [a^-3b² / ab^-15]

[2a³b⁴ / a²b³ ]² / [ab¹⁵ / a³b² ]

[4a⁶b⁸ / a⁴b⁶ ] = [4a²b² ]

and [ab¹⁵ / a³b² ] = [a¹²b¹³ ]

I am unsure about the second half.

 

[icystrike]

$$(\frac{a^{-2}b^{-3}}{2a^{3}b^{-4}})(\frac{a^{-3}b^{2}}{ab^{-15}})$$
First take a look at the left side;

It will give
$$\frac{a^{(-2-3)}b^{[-3-(-4)]}}{2}$$
Which is equal to
$$\frac{a^{-5}b}{2}$$

Now look at right side;
$$a^{(-3-1)}b^{[2-(-15)]}$$
$$a^{-4}b^{13}$$

Combine both LHS and RHS;
$$\frac{a^{[-5+(-4)]}b^{(1+1)}}{2}$$

I will leave the rest to you, noting that $$a^{-9} = \frac{1}{a^{9}}$$

 

[Sirsh]

Did you neglect to use the square of the left side? I don't see where you used it.

 

[icystrike]

Did you neglect to use the square of the left side? I don't see where you used it.

opps.. missed it.
then i guess you just have to square LHS before combining.

 

[Sirsh]

That's okay :) and btw, the LFH is divided by the RHS. sorry for confusion I don't know how to work the script stuff on this site.

 

[Sirsh]

[ $$\frac{a^-^2b^-^3}{2a^3b^-^4}$$ ]² / $$\frac{ab^-^1^5}{a^-3b^2}$$

$$\frac{a^-^4b^-^6}{4a^6b^-8}$$ / $$\frac{ab^-^1^5}{a^-^3b^2}$$

$$\frac{a^(^-^4^-^6^)b^(^-^6^-^(^-^8^)^)}{4}$$ / $$\frac{a^(^-^1^-^3^)b^(^-^1^5^-^2^)}$$

$$\frac{a^-^1^0b^2}{4}$$ / $$\frac{a^2b^-^1^7}$$

$$\frac{a^(^-^1^0^-^2^)b^(^2^-^(^-^1^7^)}{4}$$

$$\frac{a^-^1^2b^1^9}{4}$$

I ended up getting $$\frac{4}{a^1^2b^1^9}$$

If you ended up trying it, did you also get that?

 

[icystrike]

$$\frac{a^-^2b^-^3}{2a^3b^-^4}$$ ]² / $$\frac{ab^-^1^5}{a^-3b^2}$$
$$\frac{a^(^-^4^-^6^)b^(^-^6^-^(^-^8^)^)}{4}$$ / $${a^(^-^1^-^3^)b^(^-^1^5^-^2^)}$$

I see mistakes in the second expression

 

[Sirsh]

Could you point them out please, I'm not very good at indices.

 

[Sirsh]

$$\frac{b^1^9}{4a^1^2}$$

The answer is $$\frac{b^1^9}{4a^1^4}$$

So, now the only thing I need to find out is where i lost a 2.

 

[caocuong93]

Hi. I'm just a newbie but would like to give it a try.

Look for any mistakes !

This is what I got after a while working it out:

$$\left[{\frac{a^{-2}b^{-3}}{2a^3b^{-4}}\right]^2*~\frac{a^{-3}b^2}{ab^{-15}} = \frac{\left(a^{-2}b^{-3}\right)^2}{\left(2a^3b^{-4}\right)^2}\right]*~\frac{a^{-3}b^2}{ab^{-15}} = {\frac{a^{-4}b^{-6}}{4a^6b^{-8}}~*~\frac{a^{-3}b^2}{ab^{-15}} = {\frac{a^{-4-3}b^{-6+2}}{4a^{6+1}b^{-8-15}} = {\frac{a^{-7}b^{-4}}{4a^{7}b^{-23}} = {\frac{b^{-4+23}}{4a^{7+7}} = {\frac{b^{19}}{4a^{14}}$$

By the way, I wonder what class you're taking, 'cause my Pre-Cal class hasn't had this kind of stuff yet.

CN

 

[caocuong93]

Double-posted, sorry! I don't know how to delete doubled posts.

 

[Anakin_k]

Hi. I'm just a newbie but would like to give it a try.

Look for any mistakes !

This is what I got after a while working it out:

$$\left[{\frac{a^{-2}b^{-3}}{2a^3b^{-4}}\right]^2*~\frac{a^{-3}b^2}{ab^{-15}} = \frac{\left(a^{-2}b^{-3}\right)^2}{\left(2a^3b^{-4}\right)^2}\right]*~\frac{a^{-3}b^2}{ab^{-15}} = {\frac{a^{-4}b^{-6}}{4a^6b^{-8}}~*~\frac{a^{-3}b^2}{ab^{-15}} = {\frac{a^{-4-3}b^{-6+2}}{4a^{6+1}b^{-8-15}} = {\frac{a^{-7}b^{-4}}{4a^{7}b^{-23}} = {\frac{b^{-4+23}}{4a^{7+7}} = {\frac{b^{19}}{4a^{14}}$$

By the way, I wonder what class you're taking, 'cause my Pre-Cal class hasn't had this kind of stuff yet.

CN

Nope, no mistakes. I'm in Pre-Calc at the moment too, none of this stuff so far. We worked on this kind of stuff in grade 11 if memory serves me right.

 

[Sirsh]

Thanks guys for the help :) I'm from australia and I'm currently doing the Introcalculus course in year 11. I use to do the Calculus one but my study load was to much.